Tiêu đề Trigonometry Varsity Practice Sheet Solution.pdf ✅

mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ  Varsity Practice Sheet Solution 1 07 mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ Trigonometric Ratios of Associated Angles 1. sin240 Gi gvb KZ? 1 2 1 0 – 3 2 DËi: – 3 2 e ̈vL ̈v: sin240 = sin(180 + 60) = – sin60 [– KviY 3q PZzf©v‡M] = – 3 2 2. sin150 Gi gvb KZ? 1 2 – 1 2 3 2 1 3 DËi: 1 2 e ̈vL ̈v: sin150 = sin(180 – 30) = sin30 = 1 2 3. sec(– 405) = ? 1 2 – 1 2 2 – 2 DËi: 2 e ̈vL ̈v: sec(– 405) = 1 cos(– 405) = 1 cos (4  90 + 45) = 1 cos45 = 1 1 2 = 2 4. cot(– 1395) Gi gvb KZ? 1 1 2 0 2 DËi: 1 e ̈vL ̈v: cot(– 1395) = 1 – tan(1395) = 1 – tan(4  360 – 45) = 1 tan45 = 1 [⸪ 4_© PZzf©v‡M]  cot(– 1395) = 1 5. †KvbwU mwVK bq? cosec(– ) = – cosec sec(– ) = sec cos(– ) = – cos cot(– ) = – cot DËi: cos(– ) = – cos e ̈vL ̈v: cos(– ) = cos 6. hw`  = 13 6 nq, Z‡e cos Gi gvb KZ? 1 2 – 1 2 3 2 – 3 2 DËi: 3 2 e ̈vL ̈v: cos     13 6 = cos    4  2 +  6 = cos  6 = 3 2 7. cos(150 – ) = 1 2 n‡j,  Gi gvb KZ? 90 75 50 30 DËi: 90 e ̈vL ̈v: cos(150 – ) = 1 2  150 –  = 60   = 90 8. cos = – 1 2 I 450 <  < 540 n‡j  Gi gvb KZ? 300 450 480 520 DËi: 480 e ̈vL ̈v: KL‡bvB n‡e bv Option cos480 = cos (5  90 + 30) = – sin30 = – 1 2 9. tan = 5 12 Ges  my2‡KvY n‡j sin + sec(– ) Gi gvbÑ 21 156 229 156 219 156 17 32 DËi: 229 156
2  Higher Math 1st Paper Chapter-7 e ̈vL ̈v: sin + sec(– ) = sin + 1 cos = 5 13 + 13 12 = 60 + 169 156 = 229 156 12 13 5 tan = 5 12  cos = 12 13 sin = 5 13 10.  abvZ¥K I my2‡KvY Ges cos= 4 5 nq, Zvn‡j tan Gi gvb n‡eÑ 4 5 4 3 3 4 3 5 †Kv‡bvwUB bq DËi: 3 4 e ̈vL ̈v: cos= 4 5 n‡j, tan = 3 4 4 3  5 11. 4 3   A C B sin + tan = KZ? 15 29 29 15 9 5 5 9 DËi: 29 15 e ̈vL ̈v: sin = 3 5 ; tan = 4 3  sin + tan = 3 5 + 4 3 = 9 + 20 15 = 29 15 4 3  5  12.  2 <  <  Ges sin = 5 13 n‡j tan + sec(– ) cot + cosec(– ) Gi gvbÑ 3 10 2 5 2 13 5 13 DËi: 3 10 e ̈vL ̈v: sin = 5 13 n‡j, tan = – 5 12      2 <  <  gv‡b 2q PZzf©vM sec = – 13 12 cot = – 12 5 cosec = 13 5 12 5 13  tan + sec(– ) cot + cosec(– ) = – 5 12 – 13 12 – 12 5 – 13 5 = (– 5 – 13)5 (– 12 – 13)12 = 18 25  5 12 = 3 10 13. sin(– ) sec(– ) tan(270 + ) Gi gvb KZ? – 1 + tan2  – tan2  1 DËi: 1 e ̈vL ̈v: sin(– ) sec(– ) tan(270 + ) = – sinsectan(3  90 + ) = – sinsec(– cot) = sin. 1 cos . cos sin = 1 14. sin + cos = 2 n‡j 0     2 e ̈ewa‡Z  Gi gvb KZ?  2  6  4  3 DËi:  4 e ̈vL ̈v: Option Test Option sin 4 + cos  4 = 1 2 + 1 2 = 2 2 = 2 15. 2(sincos + 3) = 3 cos + 4sin Ges 0 <  < 300 n‡j  = KZ? 0, 150 0, 90 60, 120 45, 135 DËi: 60, 120 e ̈vL ̈v: 2(sincos + 3) = 3 cos + 4sin  2sincos + 2 3 – 3 cos – 4sin = 0  2sin(cos – 2) – 3 (cos – 2) = 0  (cos – 2) (2sin – 3) = 0  2sin – 3 = 0 [ cos – 2  0]  sin = 3 2   = 60, 120
mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ  Varsity Practice Sheet Solution 3 16. 8sin4 2 – 8sin2 2 + 1 Gi gvb †KvbwU? 2sin2  2cos2  cos2 sin2 DËi: cos2 e ̈vL ̈v: awi,  = 60  8sin4 30 – 8sin2 30 + 1 = 8.    1 2 4 – 8.    1 2 2 + 1 = 1 2 – 1 = – 1 2 Option I †_‡K †Kv‡bvfv‡eB (– ve) gvb Avm‡e bv| Option cos(2  60 = – 1 2 17. lnsin = 0 n‡j cos = ? 1 0 1 2 3 2 DËi: 0 e ̈vL ̈v: lnsin = 0  sin = e0 = 1  =  2  cos  2 = 0 18. hw` 3sec4  + 8 = 10sec2  nq, Z‡e tan Gi gvb n‡eÑ  1 3  1  1 2 ,  1  1 3 ,  1 DËi:  1 3 ,  1 e ̈vL ̈v: 3(1 + tan2 ) 2 + 8 – 10 – 10 tan2  = 0  3 + 6tan2  + 3tan4  –2 – 10tan2  = 0  3tan4  – 3tan2  – tan2  + 1 = 0  3tan2  (tan2  – 1) – 1 (tan2 – 1) = 0 tan 2  = 1 ; tan2  = 1 3 tan =  1 ; tan =  1 3 19. sin2      4 – A + sin2      4 + A Gi gvb †KvbwU? 0 1 – 1 1 2 DËi: 1 e ̈vL ̈v: awi, A = 0  sin2 4 + sin2 4 = 1 2 + 1 2 = 1 20. sin2 7 + sin2 5 14 + sin28 7 + sin29 14 1 2 3 4 DËi: 2 e ̈vL ̈v: sin2 7 + sin2      2 –  7 + sin2     2  2 +  7 + sin2      2 +  7 = sin2 7 + cos2 7 + sin2 7 + cos2 7 = 2 21. sin(A – 30) + sin(150 + A) Gi gvbÑ – 1 2 cosA 0 cosA sinA DËi: 0 e ̈vL ̈v: awi, A = 60  sin30 + sin210  sin30 – sin30 = 0 22. cos2 0 + cos2 10 + cos2 20 + ..... + cos2 90 Gi gvbÑ 6 3 5 4 DËi: 5 e ̈vL ̈v: cos2 0 + cos2 10 +..... + cos2 80 + cos2 90 = cos2 0 + cos2 10 + cos2 20 + cos2 30 + cos2 40 + sin2 40 + sin2 30 + sin2 20 + sin2 10 + sin2 0 = 1 + 1 + 1 + 1 + 1 = 5 A_ev: c`msL ̈v = †klc` – cÖ_gc` mvaviY AšÍi + 1 = 90 – 0 10 + 1 = 10 gvb = c`msL ̈v 2 = 10 2 = 5 23. sin2 (1 + cot2 ) + cos2 (1 + tan2 ) = ? 1 2 0 4 DËi: 2 e ̈vL ̈v: sin2 (1 + cot2 ) + cos2 (1 + tan2 ) = sin2  cosec2  + cos2  sec2   24. hw` n Gi gvb †Rvo nq, Z‡e cos + cos ( + ) + cos(2 + ) + .... + cos(n + ) = ? cos 0 ncos cosn DËi: cos e ̈vL ̈v: cos ( + ) = – cos cos(2 + ) = cos n †Rvo msL ̈v n‡j, cos(n + ) = cos cos{(n – 1)  + } = – cos cos + cos( + ) + .... + cos{(n – 1)  + } + cos(n + ) = 0 + cos
4  Higher Math 1st Paper Chapter-7 25. cot9 cot27 cot45 cot63 cot81 (0.5 + 0.5) (0.5 – 0.5) (0.25 + 0.25) – (0.25 + 0.25) DËi: (0.5 + 0.5) e ̈vL ̈v: cot9 cot27 cot45 cot(90 – 27) cot(90 – 9) = cot9tan9cot27 tan27 cot45 = 1 = (0.5 + 0.5) 26. sin2 2 + sin2 7 + sin2 12 + ..... + sin2 147 Gi c` msL ̈v KZ? 15 27 29 30 DËi: 30 e ̈vL ̈v: c`msL ̈v = †klc` – cÖ_gc` mvavib AšÍi + 1 = 147 – 2 5 + 1 = 30 27. 3  i=1 (sin2 i + cos2 i) = ? 1 3 1 + 2 + 3 †Kv‡bvwUB bq DËi: 3 e ̈vL ̈v: sin2 1 + cos2 1 = 1 Abyiƒcfv‡e i = 1 †_‡K 3 ch©šÍ †hvM Ki‡j, 1 + 1 + 1 = 3 28. tan15 + tan45 + tan75 + ..... + tan165 = ? 0 1 3 6 DËi: 0 e ̈vL ̈v: tan15 + tan45 +......+ tan(180 – 45) + tan(180 – 15) c` msL ̈v = 165 – 15 30 + 1 = 6 wU A_©vr 3 †Rvov tan15 + tan(180 – 15) = 0  tan15 + tan45 + tan75 + ..... + tan165 = 0 + 0 + 0 = 0 29. sin780cos390 – sin330 cos(– 300) = ? – 1 0 1 1 2 DËi: 1 e ̈vL ̈v: sin(8  90 + 60) cos(4  90 + 30) – sin(3  90 + 60) cos(3  90 + 30) = sin60cos30+ cos60sin30 = sin(60 + 30) = sin90 = 1 30. sin(x + y) cos(x – y) + cos(x + y) sin(x – y) Gi gvb KZ? 2sinx 2siny – sin2y sin2y sin2x DËi: sin2x e ̈vL ̈v: sin(x + y + x – y) = sin2x 31. A = 45 Ges cos(A + ) cos (A – ) = cos2 2 n‡j, tan(A + ) + tan(A – ) = KZ? 2sin2 2cos2 2sec2 2cosec2 DËi: 2sec2 e ̈vL ̈v: tan(A + ) + tan(A – ) = sin(A + ) cos(A – ) + sin(A – ) cos(A + ) cos(A + ) cos(A – ) = 2 cos2 = 2sec2 32. cos6820 cos820 + cos8140 cos2140 Gi gvb †KvbwU? 1 1 2 3 2 1 2 DËi: 1 2 e ̈vL ̈v: cos6820 cos820 + sin820 sin6820 = cos(6820 – 820) = cos60 = 1 2 33. tan(45 + A) + tan(45 – A) Gi gvb KZ? 2sinA 2cosA 2tanA 3cotA 2sec2A DËi: 2sec2A e ̈vL ̈v: awi, A = 90  tan(90 + 45) + tan(– 45) = – cot45 – tan45 = – 2 Option 2sec2A = 2sec(2  90 = 2 cos180 = – 2 [A Gi Ggb gvb ai‡Z n‡e †hb Ackb Avjv`v nq] 34. sinxsin(x + 30) + cosxsin(x + 120) Gi gvb KZ? 3 2 1 2 1 2 3 DËi: 3 2 e ̈vL ̈v: awi, x = 0  sin0sin30 + cos0sin120 = 3 2 35. sec75 Gi gvb KZ? 2 2 3 – 1 2 2 3 + 1 3 + 1 2 2 3 – 1 2 2 DËi: 2 2 3 – 1